Video Transcript
Find the equation of the normal to
the curve three 𝑦 squared minus nine 𝑦𝑥 plus seven 𝑥 squared equals one at the
point negative one, negative one.
The curve we’ve been given has been
defined implicitly. It is a function of both 𝑥 and 𝑦
and cannot be easily rearranged to a form in which we have 𝑦 as a function of
𝑥. To find the equation of the normal
to this curve, we’re going to use the point–slope form of the equation of a straight
line. That’s 𝑦 minus 𝑦 one equals 𝑚 𝑥
minus 𝑥 one. We know the coordinates of a point
on the normal, the point negative one, negative one. But we need to calculate its
slope. We recall that a normal to a curve
is perpendicular to the tangent to the curve at that point. So first, we must calculate the
slope of the tangent. To do this, we’re going to need to
find the slope function of the curve. So we need to recall how we
differentiate a function which is defined implicitly.
Implicit differentiation is an
application of the chain rule. Recall that if 𝑦 is a function of
𝑢 and 𝑢 is a function of 𝑥, then the chain rule states that d𝑦 by d𝑥 is equal
to d𝑦 by d𝑢 multiplied by d𝑢 by d𝑥. So the derivative of 𝑦 with
respect to 𝑥 is found by taking the derivative of 𝑦 with respect to 𝑢 and then
multiplying this by the derivative of 𝑢 with respect to 𝑥. Let’s consider then how we can
apply this to the equation of our curve. We’ll do this term by term,
starting with the first term three 𝑦 squared. As three 𝑦 squared is a function
of 𝑦 and we can consider 𝑦 to be a function of 𝑥, by the chain rule, we have that
the derivative with respect to 𝑥 of three 𝑦 squared is equal to the derivative
with respect to 𝑦 of three 𝑦 squared multiplied by d𝑦 by d𝑥.
By the power rule of
differentiation, the derivative with respect to 𝑦 of three 𝑦 squared is simply six
𝑦. So we find that the derivative with
respect to 𝑥 of three 𝑦 squared is six 𝑦 d𝑦 by d𝑥. So, we’ve differentiated the first
term in our equation implicitly. Differentiating the third term is
straightforward because it is a function of 𝑥 only. By the power rule of
differentiation, the derivative with respect to 𝑥 of seven 𝑥 squared is 14𝑥. And differentiating the term on the
right-hand side of the equation is straightforward because it is a constant. And we know that the derivative of
any constant with respect to 𝑥 is simply zero.
We have one term left to
differentiate, negative nine 𝑦𝑥. And this is a little more
complicated as it is a product involving both 𝑦 and 𝑥. We’ll need to use implicit
differentiation. But we also need to recall the
product rule. This states that for two
differentiable functions 𝑢 and 𝑣, the derivative with respect to 𝑥 of their
product 𝑢𝑣 is equal to 𝑢 times d𝑣 by d𝑥 plus 𝑣 times d𝑢 by d𝑥. We’ll therefore let 𝑢 equal
negative nine 𝑦 and 𝑣 equal 𝑥. We then need to find their
individual derivatives with respect to 𝑥. d𝑣 by d𝑥 is straightforward because 𝑣
is a function of 𝑥 only. If 𝑣 is equal to 𝑥, then d𝑣 by
d𝑥 is equal to one.
For d𝑢 by d𝑥 though, as 𝑢 is a
function of 𝑦, we’re going to need to use implicit differentiation again. By the chain rule, we have that the
derivative with respect to 𝑥 of negative nine 𝑦 is equal to the derivative with
respect to 𝑦 of negative nine 𝑦 multiplied by d𝑦 by d𝑥. The derivative with respect to 𝑦
of negative nine 𝑦 is simply negative nine. So we have that d𝑢 by d𝑥 is equal
to negative nine d𝑦 by d𝑥. We then need to substitute each of
these expressions into the product rule. We have that the derivative with
respect to 𝑥 of negative nine 𝑦𝑥 is equal to 𝑢 times d𝑣 by d𝑥. That’s negative nine 𝑦 multiplied
by one. Then we add 𝑣 times d𝑢 by
d𝑥. That’s 𝑥 multiplied by negative
nine d𝑦 by d𝑥. And this will simplify to negative
nine 𝑦 minus nine 𝑥 d𝑦 by d𝑥.
So we’ve now differentiated the
entire equation, giving six 𝑦 d𝑦 by d𝑥 minus nine 𝑦 minus nine 𝑥 d𝑦 by d𝑥
plus 14𝑥 is equal to zero. We want to evaluate the slope
function at a particular point. So first, we need to rearrange this
equation to give an explicit expression for d𝑦 by d𝑥. In other words, we need to make d𝑦
by d𝑥 the subject of this equation. We begin by collecting the terms
that do involve d𝑦 by d𝑥 on one side of the equation and the terms that don’t on
the other. So we have six 𝑦 d𝑦 by d𝑥 minus
nine 𝑥 d𝑦 by d𝑥 is equal to nine 𝑦 minus 14𝑥. We can then factor the left-hand
side of the equation to give six 𝑦 minus nine 𝑥 multiplied by d𝑦 by d𝑥 is equal
to nine 𝑦 minus 14𝑥.
And to find an expression for d𝑦
by d𝑥 which is in terms of both 𝑥 and 𝑦, we divide both sides of the equation by
six 𝑦 minus nine 𝑥. So we have d𝑦 by d𝑥 is equal to
nine 𝑦 minus 14𝑥 over six 𝑦 minus nine 𝑥. We have an expression for the
general slope function of the curve. But we need to evaluate the slope
at a particular point, the point negative one, negative one. We therefore need to substitute the
value negative one for both 𝑥 and 𝑦. We have negative nine plus 14 in
the numerator and negative six plus nine in the denominator, which simplifies to the
fraction five over three.
Remember though that this is the
slope of the tangent to the curve at the point negative one, negative one, whereas
we’re looking to find the equation of the normal. But we know that the slope of the
normal is the negative reciprocal of the slope of the tangent. So we have that the slope of the
normal is equal to negative three-fifths. We can invert the fraction and
change the sign.
As we now know the coordinates of a
point on the normal and its slope, we can find its equation using the point–slope
form of the equation of a straight line. We have 𝑦 minus negative one is
equal to negative three-fifths 𝑥 minus negative one. Of course, that’s just 𝑦 plus one
equals negative three-fifths 𝑥 plus one. We can multiply through by five and
distribute the parentheses on the right-hand side to give five 𝑦 plus five equals
negative three 𝑥 minus three. And finally, we’ll group all of the
terms on the left-hand side of the equation. Using implicit differentiation
then, we found that the equation of the normal to the given curve at the point
negative one, negative one is five 𝑦 plus three 𝑥 plus eight equals zero.
